What is 2-1/2 log_{10}(x)-2log_{10}(y) ?

The solution to 2-1/2 log_{10}(x)-2log_{10}(y) is 2-log_{10}(x^{1/2}y^2)

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simplify 2-1/2 log_{10}(x)-2log_{10}(y)

simplify

2 - \frac{1}{2} lo g_{10} (x) - 2 lo g_{10} (y)

2 - lo g_{10} (x^{\frac{1}{2}} y^{2})

Solution steps

2 - \frac{1}{2} lo g_{10} (x) - 2 lo g_{10} (y)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{1}{2} lo g_{10} (x) = lo g_{10} (x^{\frac{1}{2}})

= 2 - lo g_{10} (x^{\frac{1}{2}}) - 2 lo g_{10} (y)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 lo g_{10} (y) = lo g_{10} (y^{2})

= 2 - lo g_{10} (x^{\frac{1}{2}}) - lo g_{10} (y^{2})

Rewrite

- lo g_{10} (x^{\frac{1}{2}}) - lo g_{10} (y^{2})

- (lo g_{10} (x^{\frac{1}{2}}) + lo g_{10} (y^{2}))

= 2 - (lo g_{10} (x^{\frac{1}{2}}) + lo g_{10} (y^{2}))

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{10} (x^{\frac{1}{2}}) + lo g_{10} (y^{2}) = lo g_{10} (y^{2} x^{\frac{1}{2}})

= 2 - lo g_{10} (x^{\frac{1}{2}} y^{2})

\frac{3 ln ( 7 ) + 7 ln ( 2 )}{ln ( 7 ) - 2 ln ( 2 )} - 0

3 lo g_{2} (x) - 2 lo g_{2} (y) + \frac{1}{2} lo g_{2} (z)

ln (12) - ln (25)

lo g_{e} (2 e^{+ e})

ln ((y - 1) e^{y})

What is 2-1/2 log_{10}(x)-2log_{10}(y) ?
The solution to 2-1/2 log_{10}(x)-2log_{10}(y) is 2-log_{10}(x^{1/2}y^2)